Noether's Theorem is used to related the invariance under certain continuous transformations to conserved currents. A common example is that translations in spacetime correspond to the conservation of four-momentum.

In the case of angular momentum, the tensor (in special relativity) has 3 independent components for the classical angular momentum, but 3 more independent components that, as far as I know, represent Lorentz boosts. So, what conservation law corresponds to invariance under Lorentz boosts?

Yeah, shortly after posting I found this link which led me to Landau & Lifshitz (1975), pp 41-42. I didn't find their explanation totally clear, but it seems the conserved current is the four-velocity of the centre of mass?
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WarrickJul 21 '11 at 10:38

You can answer your own question @Warrick. In fact, I think there's even a badge for that ;-)
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qftmeJul 21 '11 at 10:57

2

To be precise, the conserved quantity for a system of free relativistic particles is the position (not the velocity) of the center of mass, which can be obtained by the application of the Noether theorem.
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David Bar MosheJul 21 '11 at 11:15

But L&L say that the three-vector $\underline{R}=\sum_i E_i\underline{v}_i/\sum_iE_i$ doesn't represent the spatial part of any four-vector? Is this the same as the fact that the three-vector $\underline{x}\wedge\underline{v}$ also isn't part of a four-vector?
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WarrickJul 21 '11 at 11:55

1 Answer
1

Warning: this is a long and boring derivation. If you are interested only in the result skip to the very last sentence.

Noether's theorem can be formulated in many ways. For the purposes of your question we can comfortably use the special relativistic Lagrangian formulation of a scalar field. So, suppose we are given an action
$$S[\phi] = \int {\mathcal L}(\phi(x), \partial \phi_{\mu}(x), \dots) {\rm d}^4x.$$

This object is more commonly known as stress energy tensor $T^{\mu \nu}$ and the associated conserved currents are known as momenta $p^{\nu}$.
Also, in general the conserved current is simply given by $J^{\mu} = T^{\mu \nu} a_{\nu}$.

Note that for particles we can proceed a little further since their associated momenta and angular momenta are not given by an integral. Therefore we have simply that $p^{\mu} = T^{\mu 0}$ and $M^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$. The rotation part of this (written in the form of the usual pseudovector) is
$${\mathbf L}_i = {1 \over 2}\epsilon_{ijk} M^{jk} = ({\mathbf x} \times {\mathbf p})_i$$
while for the boost part we get
$$M^{0 i} = \left(t {\mathbf p} - {\mathbf x} E \right)^i $$
which is nothing else than the center of mass at $t=0$ (we are free to choose $t$ since the quantity is conserved) multiplied by $\gamma$ since we have the relations $E = \gamma m$, ${\mathbf p} = \gamma m {\mathbf v}$. Note the similarity to the ${\mathbf E}$, $\mathbf B$ decomposition of the electromagnetic field tensor $F^{\mu \nu}$. .........